Integrand size = 31, antiderivative size = 99 \[ \int \frac {\cos ^5(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {5 x}{16 a}+\frac {i \cos ^6(c+d x)}{6 a d}+\frac {5 \cos (c+d x) \sin (c+d x)}{16 a d}+\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}+\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a d} \]
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Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3171, 3169, 2715, 8, 2645, 30} \[ \int \frac {\cos ^5(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {i \cos ^6(c+d x)}{6 a d}+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 a d}+\frac {5 \sin (c+d x) \cos ^3(c+d x)}{24 a d}+\frac {5 \sin (c+d x) \cos (c+d x)}{16 a d}+\frac {5 x}{16 a} \]
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Rule 8
Rule 30
Rule 2645
Rule 2715
Rule 3169
Rule 3171
Rubi steps \begin{align*} \text {integral}& = -\frac {i \int \cos ^5(c+d x) (i a \cos (c+d x)+a \sin (c+d x)) \, dx}{a^2} \\ & = -\frac {i \int \left (i a \cos ^6(c+d x)+a \cos ^5(c+d x) \sin (c+d x)\right ) \, dx}{a^2} \\ & = -\frac {i \int \cos ^5(c+d x) \sin (c+d x) \, dx}{a}+\frac {\int \cos ^6(c+d x) \, dx}{a} \\ & = \frac {\cos ^5(c+d x) \sin (c+d x)}{6 a d}+\frac {5 \int \cos ^4(c+d x) \, dx}{6 a}+\frac {i \text {Subst}\left (\int x^5 \, dx,x,\cos (c+d x)\right )}{a d} \\ & = \frac {i \cos ^6(c+d x)}{6 a d}+\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}+\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a d}+\frac {5 \int \cos ^2(c+d x) \, dx}{8 a} \\ & = \frac {i \cos ^6(c+d x)}{6 a d}+\frac {5 \cos (c+d x) \sin (c+d x)}{16 a d}+\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}+\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a d}+\frac {5 \int 1 \, dx}{16 a} \\ & = \frac {5 x}{16 a}+\frac {i \cos ^6(c+d x)}{6 a d}+\frac {5 \cos (c+d x) \sin (c+d x)}{16 a d}+\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}+\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a d} \\ \end{align*}
Time = 0.86 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^5(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {60 c+60 d x+15 i \cos (2 (c+d x))+6 i \cos (4 (c+d x))+i \cos (6 (c+d x))+45 \sin (2 (c+d x))+9 \sin (4 (c+d x))+\sin (6 (c+d x))}{192 a d} \]
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Time = 1.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {5 x}{16 a}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{192 a d}+\frac {i \cos \left (4 d x +4 c \right )}{32 a d}+\frac {3 \sin \left (4 d x +4 c \right )}{64 a d}+\frac {5 i \cos \left (2 d x +2 c \right )}{64 a d}+\frac {15 \sin \left (2 d x +2 c \right )}{64 a d}\) | \(96\) |
derivativedivides | \(\frac {-\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{32}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{24 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {3}{16 \left (\tan \left (d x +c \right )-i\right )}+\frac {i}{32 \left (\tan \left (d x +c \right )+i\right )^{2}}+\frac {5 i \ln \left (\tan \left (d x +c \right )+i\right )}{32}+\frac {1}{8 \tan \left (d x +c \right )+8 i}}{d a}\) | \(102\) |
default | \(\frac {-\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{32}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{24 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {3}{16 \left (\tan \left (d x +c \right )-i\right )}+\frac {i}{32 \left (\tan \left (d x +c \right )+i\right )^{2}}+\frac {5 i \ln \left (\tan \left (d x +c \right )+i\right )}{32}+\frac {1}{8 \tan \left (d x +c \right )+8 i}}{d a}\) | \(102\) |
parallelrisch | \(\frac {120 i d x \sin \left (d x +c \right )+120 d x \cos \left (d x +c \right )+16 i \cos \left (d x +c \right )-i \cos \left (5 d x +5 c \right )-15 i \cos \left (3 d x +3 c \right )+104 \sin \left (d x +c \right )+5 \sin \left (5 d x +5 c \right )+45 \sin \left (3 d x +3 c \right )}{384 a d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}\) | \(112\) |
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Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^5(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {{\left (120 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 30 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 60 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{384 \, a d} \]
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Time = 0.22 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.21 \[ \int \frac {\cos ^5(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\begin {cases} \frac {\left (- 50331648 i a^{4} d^{4} e^{16 i c} e^{4 i d x} - 503316480 i a^{4} d^{4} e^{14 i c} e^{2 i d x} + 1006632960 i a^{4} d^{4} e^{10 i c} e^{- 2 i d x} + 251658240 i a^{4} d^{4} e^{8 i c} e^{- 4 i d x} + 33554432 i a^{4} d^{4} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{6442450944 a^{5} d^{5}} & \text {for}\: a^{5} d^{5} e^{12 i c} \neq 0 \\x \left (\frac {\left (e^{10 i c} + 5 e^{8 i c} + 10 e^{6 i c} + 10 e^{4 i c} + 5 e^{2 i c} + 1\right ) e^{- 6 i c}}{32 a} - \frac {5}{16 a}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{16 a} \]
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Exception generated. \[ \int \frac {\cos ^5(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^5(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {-\frac {30 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a} + \frac {30 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {3 \, {\left (-15 i \, \tan \left (d x + c\right )^{2} + 38 \, \tan \left (d x + c\right ) + 25 i\right )}}{a {\left (-i \, \tan \left (d x + c\right ) + 1\right )}^{2}} - \frac {55 i \, \tan \left (d x + c\right )^{3} + 201 \, \tan \left (d x + c\right )^{2} - 255 i \, \tan \left (d x + c\right ) - 117}{a {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{192 \, d} \]
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Time = 29.37 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.66 \[ \int \frac {\cos ^5(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {5\,x}{16\,a}+\frac {\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,3{}\mathrm {i}}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,1{}\mathrm {i}}{12}+\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,1{}\mathrm {i}}{12}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}}{4}+\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^4\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^6} \]
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